{"id":436,"date":"2025-02-13T19:44:54","date_gmt":"2025-02-13T19:44:54","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/determining-volumes-by-slicing-fresh-take\/"},"modified":"2025-02-13T19:44:54","modified_gmt":"2025-02-13T19:44:54","slug":"determining-volumes-by-slicing-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/determining-volumes-by-slicing-fresh-take\/","title":{"raw":"Determining Volumes by Slicing: Fresh Take","rendered":"Determining Volumes by Slicing: Fresh Take"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Find the volume of a solid by using the slicing method<\/li>\n\t<li>Find the volume of a solid by using the disk method<\/li>\n\t<li>Compute the volume of a hollow solid of revolution by using the washer technique<\/li>\n<\/ul>\n<\/section>\n<h2>Volume and the Slicing Method<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Volume is a measure of three-dimensional space occupied by a solid. While we have formulas for basic shapes (e.g., [latex]V = \\frac{4}{3}\\pi r^3[\/latex] for a sphere), not all solids have simple formulas.<\/li>\n\t<li class=\"whitespace-normal break-words\">The slicing method is a technique for finding volumes of solids with varying cross-sections, extending the concept of definite integrals to three dimensions.<\/li>\n\t<li class=\"whitespace-normal break-words\">Key formula:\n\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]V = \\int_a^b A(x) dx[\/latex], where [latex]A(x)[\/latex] is the cross-sectional area<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">A cylinder, in mathematical terms, is any solid generated by translating a plane region along a line perpendicular to it (the axis). This generalizes the common notion of a circular cylinder.<\/li>\n\t<li class=\"whitespace-normal break-words\">Cross-sections are two-dimensional slices of a three-dimensional solid, obtained by intersecting the solid with a plane. The shape and area of these cross-sections are crucial in applying the slicing method.<\/li>\n\t<li class=\"whitespace-normal break-words\">The choice of slicing direction (e.g., perpendicular to [latex]x[\/latex]-axis, [latex]y[\/latex]-axis, or [latex]z[\/latex]-axis) can significantly affect the complexity of calculations. Selecting the appropriate direction is a key problem-solving skill.<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Slicing Method Process<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Divide the solid into thin slices perpendicular to a chosen axis<\/li>\n\t<li class=\"whitespace-normal break-words\">Estimate volume of each slice: [latex]V(S_i) \\approx A(x_i^*) \\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Sum the volumes of all slices: [latex]V(S) \\approx \\sum_{i=1}^n A(x_i^*) \\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Take the limit as[latex] n[\/latex] approaches infinity to get the definite integral<\/li>\n<\/ol>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Determine the shape of a cross-section of the solid<\/li>\n\t<li class=\"whitespace-normal break-words\">Find a formula for the area of the cross-section<\/li>\n\t<li class=\"whitespace-normal break-words\">Integrate the area formula over the appropriate interval<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\">\n<p>Use the slicing method to derive the formula [latex]V=\\frac{1}{3}\\pi {r}^{2}h[\/latex] for the volume of a circular cone.<\/p>\n\n\n[reveal-answer q=\"17667\"]Show Answer[\/reveal-answer] [hidden-answer a=\"17667\"]\n\n\n<p>Consider a cone with height [latex]h[\/latex] and base radius [latex]r[\/latex]. The cone can be viewed as being composed of an infinite number of thin disks stacked along its height.<\/p>\n<p>We slice the cone horizontally at a height [latex]y[\/latex] from the base. At this height, the radius of the disk [latex]r(y)[\/latex] varies linearly from [latex]0[\/latex] at the top to [latex]r[\/latex] at the base.<\/p>\n<p>The relationship between [latex]r(y)[\/latex] and [latex]y[\/latex] can be expressed using similar triangles.<\/p>\n<p>Since the height of the cone is [latex]h[\/latex] and the radius at the base is [latex]r[\/latex], we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{r(y)}{y} = \\frac{r}{h} \\implies r(y) = \\frac{r}{h} y[\/latex]<\/p>\n<p>Each thin disk has a thickness [latex]dy[\/latex] and a radius [latex]r(y)[\/latex].<\/p>\n<p>The volume [latex]dV[\/latex] of a thin disk at height [latex]y[\/latex] is the area of the disk times its thickness:<\/p>\n<p style=\"text-align: center;\">[latex]dV = \\pi [r(y)]^2 dy[\/latex]<\/p>\n<p>Substitute [latex]r(y) = \\frac{r}{h} y[\/latex] into the volume element expression:<\/p>\n<p style=\"text-align: center;\">[latex]dV = \\pi \\left( \\frac{r}{h} y \\right)^2 dy = \\pi \\frac{r^2}{h^2} y^2 dy[\/latex]<\/p>\n<p>To find the total volume of the cone, integrate [latex]dV[\/latex] from the base ([latex]y = 0[\/latex]) to the top ([latex]y = h[\/latex]):<\/p>\n<p style=\"text-align: center;\">[latex]V = \\int_0^h \\pi \\frac{r^2}{h^2} y^2 dy[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\pi \\frac{r^2}{h^2} \\int_0^h y^2 dy[\/latex]<\/p>\n<p>The integral of [latex]y^2[\/latex] with respect to [latex]y[\/latex] is:<\/p>\n<p style=\"text-align: center;\">[latex]\\int y^2 dy = \\frac{y^3}{3}[\/latex] So, we have: [latex]V = \\pi \\frac{r^2}{h^2} \\left[ \\frac{y^3}{3} \\right]_0^h = \\pi \\frac{r^2}{h^2} \\left( \\frac{h^3}{3} - 0 \\right) = \\pi \\frac{r^2}{h^2} \\cdot \\frac{h^3}{3} = \\frac{1}{3} \\pi r^2 h[\/latex]<\/p>\n<p style=\"text-align: left;\">Thus, we have derived the formula for the volume of a circular cone:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\frac{1}{3} \\pi r^2 h[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find the volume of a right circular cone with radius [latex]r[\/latex] and height [latex]h[\/latex] using the slicing method.<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><br>\n[reveal-answer q=\"367772\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"367772\"]<\/p>\n<p>Visualize the cone with its base on the [latex]xy[\/latex]-plane and apex at [latex](0, 0, h)[\/latex]. Cross-sections perpendicular to the [latex]z[\/latex]-axis are circles<\/p>\n<p><br>\nFind the radius of a circular cross-section at height [latex]z[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{r}{h} = \\frac{r - R}{h - z}[\/latex], where [latex]R[\/latex] is the radius at height [latex]z[\/latex]<\/p>\n<p>Solving for R:<\/p>\n<p style=\"text-align: center;\">[latex]R = r - \\frac{rz}{h}[\/latex]<\/p>\n<p>Area of the circular cross-section:<\/p>\n<p style=\"text-align: center;\">[latex]A(z) = \\pi R^2 = \\pi (r - \\frac{rz}{h})^2[\/latex]<\/p>\n<p>Set up the integral:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\int_0^h A(z) dz = \\int_0^h \\pi (r - \\frac{rz}{h})^2 dz[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\nV &amp;=&amp; \\pi \\int_0^h (r^2 - \\frac{2r^2z}{h} + \\frac{r^2z^2}{h^2}) dz \\\\<br>\n&amp;=&amp; \\pi [r^2z - \\frac{r^2z^2}{h} + \\frac{r^2z^3}{3h^2}]_0^h \\\\<br>\n&amp;=&amp; \\pi (r^2h - \\frac{r^2h}{1} + \\frac{r^2h}{3}) \\\\<br>\n&amp;=&amp; \\frac{1}{3}\\pi r^2h<br>\n\\end{array}[\/latex]<\/p>\n<p>This result matches the known formula for the volume of a cone, validating our use of the slicing method.<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\n<\/section>\n<p>&nbsp;<\/p>\n<h2>Solids of Revolution and the Slicing Method<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">A solid of revolution is formed by rotating a planar region around a line in that plane<\/li>\n\t<li class=\"whitespace-normal break-words\">Cross-sections of these solids perpendicular to the axis of revolution are typically circular<\/li>\n\t<li class=\"whitespace-normal break-words\">Volume formula:\n\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]V = \\int_a^b A(x) dx[\/latex], where [latex]A(x)[\/latex] is the area of the circular cross-section<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">For rotation around [latex]x[\/latex]-axis:\n\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]A(x) = \\pi [f(x)]^2[\/latex], where [latex]f(x)[\/latex] is the function being rotated<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">The choice of axis of revolution affects the complexity of the integration<\/li>\n\t<li class=\"whitespace-normal break-words\">Solids of revolution are common in manufacturing and engineering applications<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Sketch the region being rotated and visualize the resulting solid<\/li>\n\t<li class=\"whitespace-normal break-words\">Identify the axis of rotation and the limits of integration<\/li>\n\t<li class=\"whitespace-normal break-words\">Determine the formula for the cross-sectional area [latex]A(x)[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Set up and evaluate the integral [latex]V = \\int_a^b A(x) dx[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the volume of the solid formed by rotating [latex]y = 4 - x^2[\/latex] from [latex]x = 0[\/latex] to [latex]x = 2[\/latex] around the [latex]x[\/latex]-axis.&nbsp;<\/p>\n<p><br>\n[reveal-answer q=\"372283\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"372283\"]<\/p>\n<p>Cross-sectional area:<\/p>\n<p style=\"text-align: center;\">[latex]A(x) = \\pi [f(x)]^2 = \\pi (4 - x^2)^2[\/latex]<\/p>\n<p>Set up the integral:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\int_0^2 \\pi (4 - x^2)^2 dx[\/latex]<\/p>\n<p>Evaluate:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\nV &amp;=&amp; \\pi \\int_0^2 (16 - 8x^2 + x^4) dx \\\\<br>\n&amp;=&amp; \\pi [16x - \\dfrac{8x^3}{3} + \\dfrac{x^5}{5}]_0^2 \\\\<br>\n&amp;=&amp; \\pi (32 - \\dfrac{64}{3} + \\dfrac{32}{5}) \\\\<br>\n&amp;=&amp; \\frac{256\\pi}{15}<br>\n\\end{array}[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the volume of the solid formed by rotating [latex]y = x + 1[\/latex] from [latex]x = 1[\/latex] to [latex]x = 3[\/latex] around the [latex]x[\/latex]-axis.<\/p>\n<p><br>\n[reveal-answer q=\"831238\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"831238\"]<\/p>\n<p>Cross-sectional area:<\/p>\n<p style=\"text-align: center;\">[latex]A(x) = \\pi [f(x)]^2 = \\pi (x + 1)^2[\/latex]<\/p>\n<p>Set up the integral:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\int_1^3 \\pi (x + 1)^2 dx[\/latex]<\/p>\n<p>Evaluate:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\nV &amp;=&amp; \\pi \\int_1^3 (x^2 + 2x + 1) dx \\\\<br>\n&amp;=&amp; \\pi [\\frac{x^3}{3} + x^2 + x]_1^3 \\\\<br>\n&amp;=&amp; \\pi (9 + 9 + 3 - \\frac{1}{3} - 1 - 1) \\\\<br>\n&amp;=&amp; \\frac{56\\pi}{3}<br>\n\\end{array}[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the volume of the solid formed by rotating [latex]y = \\frac{1}{x}[\/latex] from [latex]x = 1[\/latex] to [latex]x = \\infty[\/latex] around the [latex]x[\/latex]-axis.<\/p>\n<p><br>\n[reveal-answer q=\"217741\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"217741\"]<\/p>\n<p>Cross-sectional area:<\/p>\n<p style=\"text-align: center;\">[latex]A(x) = \\pi [f(x)]^2 = \\pi (\\frac{1}{x})^2[\/latex]<\/p>\n<p>Set up the integral: [latex]V = \\int_1^\\infty \\pi (\\frac{1}{x})^2 dx[\/latex]<\/p>\n<p>Evaluate:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\nV &amp;=&amp; \\pi \\int_1^\\infty \\frac{1}{x^2} dx \\\\<br>\n&amp;=&amp; \\pi [-\\frac{1}{x}]_1^\\infty \\\\<br>\n&amp;=&amp; \\pi (0 - (-1)) \\\\<br>\n&amp;=&amp; \\pi<br>\n\\end{array}[\/latex]<\/p>\n<p>This example, known as Gabriel's Horn, shows a solid with finite volume but infinite surface area.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<h2>The Disk Method<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">The Disk Method is a specialized application of the slicing method for solids of revolution<\/li>\n\t<li class=\"whitespace-normal break-words\">It uses circular disks to approximate the volume of the solid<\/li>\n\t<li class=\"whitespace-normal break-words\">Key formula for rotation around [latex]x[\/latex]-axis:\n\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]V = \\int_a^b \\pi [f(x)]^2 dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Key formula for rotation around [latex]y[\/latex]-axis:\n\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]V = \\int_c^d \\pi [g(y)]^2 dy[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Applies to solids formed by rotating a region bounded by a function and an axis<\/li>\n\t<li class=\"whitespace-normal break-words\">Can be used for rotation around any horizontal or vertical line, not just coordinate axes<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Identify the function and the axis of rotation<\/li>\n\t<li class=\"whitespace-normal break-words\">Determine the limits of integration (the interval over which the region is rotated)<\/li>\n\t<li class=\"whitespace-normal break-words\">Set up the integral using the appropriate formula<\/li>\n\t<li class=\"whitespace-normal break-words\">Evaluate the integral to find the volume<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\">Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of [latex]f(x)=\\sqrt{4-x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,4\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex] [reveal-answer q=\"1167794039169\"]Show Solution[\/reveal-answer] [hidden-answer a=\"1167794039169\"]\n\n\n<p id=\"fs-id1167794039169\">[latex]8\\pi [\/latex] units<sup>3<\/sup><\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/7dG21yfKXHg?controls=0&amp;start=755&amp;end=847&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.2DeterminingVolumesBySlicing755to847_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.2 Determining Volumes by Slicing\" here (opens in new window)<\/a>.<\/p>\n\n\n[hidden-answer]<\/section>\n<h2>The Washer Method<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">The Washer Method is used for solids of revolution with cavities<\/li>\n\t<li class=\"whitespace-normal break-words\">Applies when the region being revolved is bounded by two functions or when the axis of revolution is not the boundary of the region<\/li>\n\t<li class=\"whitespace-normal break-words\">Key formula for rotation around [latex]x[\/latex]-axis:\n\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]V = \\int_a^b \\pi [(f(x))^2 - (g(x))^2] dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Key formula for rotation around [latex]y[\/latex]-axis:\n\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]V = \\int_c^d \\pi [(u(y))^2 - (v(y))^2] dy[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">The cross-sectional area is the difference between the areas of two circles<\/li>\n\t<li class=\"whitespace-normal break-words\">Can be applied to rotation around any horizontal or vertical line, not just coordinate axes<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Identify the functions bounding the region and the axis of rotation<\/li>\n\t<li class=\"whitespace-normal break-words\">Determine the radii of the outer and inner circles at each cross-section<\/li>\n\t<li class=\"whitespace-normal break-words\">Set up the integral using the appropriate washer formula<\/li>\n\t<li class=\"whitespace-normal break-words\">Evaluate the integral to find the volume<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\">\n<p>Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of [latex]f(x)=x+2[\/latex] and below by the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,3\\right][\/latex] around the line [latex]y=-1.[\/latex]<\/p>\n\n\n[reveal-answer q=\"fs-id1167793638825\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793638825\"]\n\n\n<p id=\"fs-id1167793638825\">[latex]60\\pi [\/latex] units<sup>3<\/sup><\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Find the volume of a solid by using the slicing method<\/li>\n<li>Find the volume of a solid by using the disk method<\/li>\n<li>Compute the volume of a hollow solid of revolution by using the washer technique<\/li>\n<\/ul>\n<\/section>\n<h2>Volume and the Slicing Method<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Volume is a measure of three-dimensional space occupied by a solid. While we have formulas for basic shapes (e.g., [latex]V = \\frac{4}{3}\\pi r^3[\/latex] for a sphere), not all solids have simple formulas.<\/li>\n<li class=\"whitespace-normal break-words\">The slicing method is a technique for finding volumes of solids with varying cross-sections, extending the concept of definite integrals to three dimensions.<\/li>\n<li class=\"whitespace-normal break-words\">Key formula:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]V = \\int_a^b A(x) dx[\/latex], where [latex]A(x)[\/latex] is the cross-sectional area<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">A cylinder, in mathematical terms, is any solid generated by translating a plane region along a line perpendicular to it (the axis). This generalizes the common notion of a circular cylinder.<\/li>\n<li class=\"whitespace-normal break-words\">Cross-sections are two-dimensional slices of a three-dimensional solid, obtained by intersecting the solid with a plane. The shape and area of these cross-sections are crucial in applying the slicing method.<\/li>\n<li class=\"whitespace-normal break-words\">The choice of slicing direction (e.g., perpendicular to [latex]x[\/latex]-axis, [latex]y[\/latex]-axis, or [latex]z[\/latex]-axis) can significantly affect the complexity of calculations. Selecting the appropriate direction is a key problem-solving skill.<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Slicing Method Process<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Divide the solid into thin slices perpendicular to a chosen axis<\/li>\n<li class=\"whitespace-normal break-words\">Estimate volume of each slice: [latex]V(S_i) \\approx A(x_i^*) \\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Sum the volumes of all slices: [latex]V(S) \\approx \\sum_{i=1}^n A(x_i^*) \\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Take the limit as[latex]n[\/latex] approaches infinity to get the definite integral<\/li>\n<\/ol>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Determine the shape of a cross-section of the solid<\/li>\n<li class=\"whitespace-normal break-words\">Find a formula for the area of the cross-section<\/li>\n<li class=\"whitespace-normal break-words\">Integrate the area formula over the appropriate interval<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\">\n<p>Use the slicing method to derive the formula [latex]V=\\frac{1}{3}\\pi {r}^{2}h[\/latex] for the volume of a circular cone.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q17667\">Show Answer<\/button> <\/p>\n<div id=\"q17667\" class=\"hidden-answer\" style=\"display: none\">\n<p>Consider a cone with height [latex]h[\/latex] and base radius [latex]r[\/latex]. The cone can be viewed as being composed of an infinite number of thin disks stacked along its height.<\/p>\n<p>We slice the cone horizontally at a height [latex]y[\/latex] from the base. At this height, the radius of the disk [latex]r(y)[\/latex] varies linearly from [latex]0[\/latex] at the top to [latex]r[\/latex] at the base.<\/p>\n<p>The relationship between [latex]r(y)[\/latex] and [latex]y[\/latex] can be expressed using similar triangles.<\/p>\n<p>Since the height of the cone is [latex]h[\/latex] and the radius at the base is [latex]r[\/latex], we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{r(y)}{y} = \\frac{r}{h} \\implies r(y) = \\frac{r}{h} y[\/latex]<\/p>\n<p>Each thin disk has a thickness [latex]dy[\/latex] and a radius [latex]r(y)[\/latex].<\/p>\n<p>The volume [latex]dV[\/latex] of a thin disk at height [latex]y[\/latex] is the area of the disk times its thickness:<\/p>\n<p style=\"text-align: center;\">[latex]dV = \\pi [r(y)]^2 dy[\/latex]<\/p>\n<p>Substitute [latex]r(y) = \\frac{r}{h} y[\/latex] into the volume element expression:<\/p>\n<p style=\"text-align: center;\">[latex]dV = \\pi \\left( \\frac{r}{h} y \\right)^2 dy = \\pi \\frac{r^2}{h^2} y^2 dy[\/latex]<\/p>\n<p>To find the total volume of the cone, integrate [latex]dV[\/latex] from the base ([latex]y = 0[\/latex]) to the top ([latex]y = h[\/latex]):<\/p>\n<p style=\"text-align: center;\">[latex]V = \\int_0^h \\pi \\frac{r^2}{h^2} y^2 dy[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\pi \\frac{r^2}{h^2} \\int_0^h y^2 dy[\/latex]<\/p>\n<p>The integral of [latex]y^2[\/latex] with respect to [latex]y[\/latex] is:<\/p>\n<p style=\"text-align: center;\">[latex]\\int y^2 dy = \\frac{y^3}{3}[\/latex] So, we have: [latex]V = \\pi \\frac{r^2}{h^2} \\left[ \\frac{y^3}{3} \\right]_0^h = \\pi \\frac{r^2}{h^2} \\left( \\frac{h^3}{3} - 0 \\right) = \\pi \\frac{r^2}{h^2} \\cdot \\frac{h^3}{3} = \\frac{1}{3} \\pi r^2 h[\/latex]<\/p>\n<p style=\"text-align: left;\">Thus, we have derived the formula for the volume of a circular cone:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\frac{1}{3} \\pi r^2 h[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find the volume of a right circular cone with radius [latex]r[\/latex] and height [latex]h[\/latex] using the slicing method.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q367772\">Show Answer<\/button><\/p>\n<div id=\"q367772\" class=\"hidden-answer\" style=\"display: none\">\n<p>Visualize the cone with its base on the [latex]xy[\/latex]-plane and apex at [latex](0, 0, h)[\/latex]. Cross-sections perpendicular to the [latex]z[\/latex]-axis are circles<\/p>\n<p>\nFind the radius of a circular cross-section at height [latex]z[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{r}{h} = \\frac{r - R}{h - z}[\/latex], where [latex]R[\/latex] is the radius at height [latex]z[\/latex]<\/p>\n<p>Solving for R:<\/p>\n<p style=\"text-align: center;\">[latex]R = r - \\frac{rz}{h}[\/latex]<\/p>\n<p>Area of the circular cross-section:<\/p>\n<p style=\"text-align: center;\">[latex]A(z) = \\pi R^2 = \\pi (r - \\frac{rz}{h})^2[\/latex]<\/p>\n<p>Set up the integral:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\int_0^h A(z) dz = \\int_0^h \\pi (r - \\frac{rz}{h})^2 dz[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> V &=& \\pi \\int_0^h (r^2 - \\frac{2r^2z}{h} + \\frac{r^2z^2}{h^2}) dz \\\\<br \/> &=& \\pi [r^2z - \\frac{r^2z^2}{h} + \\frac{r^2z^3}{3h^2}]_0^h \\\\<br \/> &=& \\pi (r^2h - \\frac{r^2h}{1} + \\frac{r^2h}{3}) \\\\<br \/> &=& \\frac{1}{3}\\pi r^2h<br \/> \\end{array}[\/latex]<\/p>\n<p>This result matches the known formula for the volume of a cone, validating our use of the slicing method.<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n<p>&nbsp;<\/p>\n<h2>Solids of Revolution and the Slicing Method<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">A solid of revolution is formed by rotating a planar region around a line in that plane<\/li>\n<li class=\"whitespace-normal break-words\">Cross-sections of these solids perpendicular to the axis of revolution are typically circular<\/li>\n<li class=\"whitespace-normal break-words\">Volume formula:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]V = \\int_a^b A(x) dx[\/latex], where [latex]A(x)[\/latex] is the area of the circular cross-section<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">For rotation around [latex]x[\/latex]-axis:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]A(x) = \\pi [f(x)]^2[\/latex], where [latex]f(x)[\/latex] is the function being rotated<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">The choice of axis of revolution affects the complexity of the integration<\/li>\n<li class=\"whitespace-normal break-words\">Solids of revolution are common in manufacturing and engineering applications<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Sketch the region being rotated and visualize the resulting solid<\/li>\n<li class=\"whitespace-normal break-words\">Identify the axis of rotation and the limits of integration<\/li>\n<li class=\"whitespace-normal break-words\">Determine the formula for the cross-sectional area [latex]A(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Set up and evaluate the integral [latex]V = \\int_a^b A(x) dx[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the volume of the solid formed by rotating [latex]y = 4 - x^2[\/latex] from [latex]x = 0[\/latex] to [latex]x = 2[\/latex] around the [latex]x[\/latex]-axis.&nbsp;<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q372283\">Show Answer<\/button><\/p>\n<div id=\"q372283\" class=\"hidden-answer\" style=\"display: none\">\n<p>Cross-sectional area:<\/p>\n<p style=\"text-align: center;\">[latex]A(x) = \\pi [f(x)]^2 = \\pi (4 - x^2)^2[\/latex]<\/p>\n<p>Set up the integral:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\int_0^2 \\pi (4 - x^2)^2 dx[\/latex]<\/p>\n<p>Evaluate:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> V &=& \\pi \\int_0^2 (16 - 8x^2 + x^4) dx \\\\<br \/> &=& \\pi [16x - \\dfrac{8x^3}{3} + \\dfrac{x^5}{5}]_0^2 \\\\<br \/> &=& \\pi (32 - \\dfrac{64}{3} + \\dfrac{32}{5}) \\\\<br \/> &=& \\frac{256\\pi}{15}<br \/> \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the volume of the solid formed by rotating [latex]y = x + 1[\/latex] from [latex]x = 1[\/latex] to [latex]x = 3[\/latex] around the [latex]x[\/latex]-axis.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q831238\">Show Answer<\/button><\/p>\n<div id=\"q831238\" class=\"hidden-answer\" style=\"display: none\">\n<p>Cross-sectional area:<\/p>\n<p style=\"text-align: center;\">[latex]A(x) = \\pi [f(x)]^2 = \\pi (x + 1)^2[\/latex]<\/p>\n<p>Set up the integral:<\/p>\n<p style=\"text-align: center;\">[latex]V = \\int_1^3 \\pi (x + 1)^2 dx[\/latex]<\/p>\n<p>Evaluate:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> V &=& \\pi \\int_1^3 (x^2 + 2x + 1) dx \\\\<br \/> &=& \\pi [\\frac{x^3}{3} + x^2 + x]_1^3 \\\\<br \/> &=& \\pi (9 + 9 + 3 - \\frac{1}{3} - 1 - 1) \\\\<br \/> &=& \\frac{56\\pi}{3}<br \/> \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the volume of the solid formed by rotating [latex]y = \\frac{1}{x}[\/latex] from [latex]x = 1[\/latex] to [latex]x = \\infty[\/latex] around the [latex]x[\/latex]-axis.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q217741\">Show Answer<\/button><\/p>\n<div id=\"q217741\" class=\"hidden-answer\" style=\"display: none\">\n<p>Cross-sectional area:<\/p>\n<p style=\"text-align: center;\">[latex]A(x) = \\pi [f(x)]^2 = \\pi (\\frac{1}{x})^2[\/latex]<\/p>\n<p>Set up the integral: [latex]V = \\int_1^\\infty \\pi (\\frac{1}{x})^2 dx[\/latex]<\/p>\n<p>Evaluate:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> V &=& \\pi \\int_1^\\infty \\frac{1}{x^2} dx \\\\<br \/> &=& \\pi [-\\frac{1}{x}]_1^\\infty \\\\<br \/> &=& \\pi (0 - (-1)) \\\\<br \/> &=& \\pi<br \/> \\end{array}[\/latex]<\/p>\n<p>This example, known as Gabriel&#8217;s Horn, shows a solid with finite volume but infinite surface area.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>The Disk Method<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">The Disk Method is a specialized application of the slicing method for solids of revolution<\/li>\n<li class=\"whitespace-normal break-words\">It uses circular disks to approximate the volume of the solid<\/li>\n<li class=\"whitespace-normal break-words\">Key formula for rotation around [latex]x[\/latex]-axis:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]V = \\int_a^b \\pi [f(x)]^2 dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key formula for rotation around [latex]y[\/latex]-axis:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]V = \\int_c^d \\pi [g(y)]^2 dy[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Applies to solids formed by rotating a region bounded by a function and an axis<\/li>\n<li class=\"whitespace-normal break-words\">Can be used for rotation around any horizontal or vertical line, not just coordinate axes<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Identify the function and the axis of rotation<\/li>\n<li class=\"whitespace-normal break-words\">Determine the limits of integration (the interval over which the region is rotated)<\/li>\n<li class=\"whitespace-normal break-words\">Set up the integral using the appropriate formula<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate the integral to find the volume<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\">Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of [latex]f(x)=\\sqrt{4-x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,4\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex] <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q1167794039169\">Show Solution<\/button> <\/p>\n<div id=\"q1167794039169\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794039169\">[latex]8\\pi[\/latex] units<sup>3<\/sup><\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/7dG21yfKXHg?controls=0&amp;start=755&amp;end=847&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.2DeterminingVolumesBySlicing755to847_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.2 Determining Volumes by Slicing&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"qdefault 1\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<\/div>\n<\/section>\n<h2>The Washer Method<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">The Washer Method is used for solids of revolution with cavities<\/li>\n<li class=\"whitespace-normal break-words\">Applies when the region being revolved is bounded by two functions or when the axis of revolution is not the boundary of the region<\/li>\n<li class=\"whitespace-normal break-words\">Key formula for rotation around [latex]x[\/latex]-axis:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]V = \\int_a^b \\pi [(f(x))^2 - (g(x))^2] dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key formula for rotation around [latex]y[\/latex]-axis:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]V = \\int_c^d \\pi [(u(y))^2 - (v(y))^2] dy[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">The cross-sectional area is the difference between the areas of two circles<\/li>\n<li class=\"whitespace-normal break-words\">Can be applied to rotation around any horizontal or vertical line, not just coordinate axes<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Identify the functions bounding the region and the axis of rotation<\/li>\n<li class=\"whitespace-normal break-words\">Determine the radii of the outer and inner circles at each cross-section<\/li>\n<li class=\"whitespace-normal break-words\">Set up the integral using the appropriate washer formula<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate the integral to find the volume<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\">\n<p>Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of [latex]f(x)=x+2[\/latex] and below by the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,3\\right][\/latex] around the line [latex]y=-1.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793638825\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793638825\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<\/div>\n<p id=\"fs-id1167793638825\">[latex]60\\pi[\/latex] units<sup>3<\/sup><\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":15,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":421,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/436"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/436\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/421"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/436\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=436"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=436"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=436"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=436"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}